Question #276794

A crate of mass 10kg is pulled up arough incline with an initial speed of 1.5m/s. the pulling force is 100N parallel to the incline which makes an angle of 30° with the horizontal. the coefficient of kinetic friction is 0.4 and the crate is pulled 5m. How much mechanical energy is lost due to friction?

Expert's answer

Energy loss is simply Kinetic friction multiply by the distance traveled. To find kinetic friction, take coefficient of friction multiply by normal force acting on the block.

N=10×9.8×cos30=84.87N×k=84.87×0.4=33.95Energy  loss=33.95×5=169.74  JN = 10 \times 9.8 \times cos30 = 84.87 \\ N \times k = 84.87 \times 0.4 = 33.95 \\ Energy\; loss = 33.95 \times 5 = 169.74 \;J


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

LATEST TUTORIALS
APPROVED BY CLIENTS